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anonymous
 5 years ago
Find the vertices, foci, and eccentricity of the ellipse.
5x^2+y^2=6
can someone help me with this?
anonymous
 5 years ago
Find the vertices, foci, and eccentricity of the ellipse. 5x^2+y^2=6 can someone help me with this?

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when x = 0, y = +sqrt(6) when y = 0, x = +sqrt(6/5) the rest of it is greek to me at the moment :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it appears to be taller than wider

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0those might be the vertices :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get how to do the problem. It should look like x^2/6/5 + y^2/30 = 1. I just dont know how to clear the fraction under the x^2 term.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.01/(6/5) just means find the reciprocal of 6/5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the equation must stay equal to 1.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0foci are c^2 = a^2  b^2....

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0[(xh)^2]/a+[(yk)^2}/b=1 what is this form?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.05^2  1^2 = 24.... is the slant = sqrt(24)? cant recall to much about it

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0i remember writing ellipses in that form

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0do the foci need to be on the x axis?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes because it is a horizontal equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0heres some information on horizontal ellipses is x^2/a^2 + y^2/b^2 = 1 a> b > 0 vertices (+ or a,0) foci (+ or  c,0)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know everything just not how to clear that 6/5 without changing the = 1 part.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if i could just put 5x^2+y^2=6 in the x^2/a^2 + y^2/b^2 = 1 form. i could solve it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would just leave the 6/5 where it is. But I think there is a 6 under the y^2, not 30.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What do you mean "solve"? You say you know how to do it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean find all the information from it and then graph it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, the vertices are (+sqrt(6/5), 0) and (0, +sqrt(6)). The foci you find by taking c^2 = a^2  b^2. So, c^2 = 6  6/5 = 24/5. c = +sqrt(24/5). The foci are on the yaxis because the ellipse is vertically oriented: (0, +sqrt(24/5)). The eccentricity you find by taking c / a, so sqrt(24/5) / sqrt(6) = sqrt(4/5).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Cool. thats what i got. Thanx!
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